#A67 INTEGERS 20 (2020) BERNOULLI-STIRLING NUMBERS

Ren´eGy [email protected]

Received: 5/29/19, Revised: 2/29/20, Accepted: 8/13/20, Published: 8/31/20

Abstract Congruences modulo prime powers involving generalized Harmonic numbers are known. While looking for similar congruences, we have encountered a curious tri- angular array of numbers indexed with positive integers n, k, involving the Bernoulli and cycle Stirling numbers. These numbers are all integers and they vanish when n − k is odd. This triangle has many similarities with the Stirling triangle. In particular, we show how it can be extended to negative indices and how this ex- tension produces a second kind of such integers which may be considered as a new generalization of the Genocchi numbers and for which a generating function is easily obtained. But our knowledge of these integers remains limited, especially for those of the first kind.

1. Introduction

Let n and k be non-negative integers and let the generalized Harmonic numbers (k) (k) Hn , Gn be defined as n X 1 H(k) := , n jk j=1

(k) X 1 Gn := , i1i2 ··· ik 1≤i1

(1) (1) Pn 1 (0) (0) with Hn = Gn = j=1 j = Hn = Gn; Hn = n and Gn = 1. We have [10] n + 1 = n!G(k), k + 1 n n k being the cycle Stirling number (or unsigned Stirling number of the first kind), so that the Harmonic and cycle Stirling numbers are inter-related by the convolution

k−1 n + 1 X n + 1 k = − (−1)k−jH(k−j) , (1.1) k + 1 n j + 1 j=0 2 which is obtained as a direct application of the well-known relation between ele- mentary symmetric polynomials and power sums [9]. Extended congruences for the (k) Harmonic numbers Hp−1, modulo any power of a prime p are known [16], [7]. Our initial motivation for the work reported in the present paper is to look for similar (k) congruences modulo prime powers, involving Gp−1, or the cycle Stirling numbers  p  (k) (k) k+1 , instead of Hp−1. We will show that such similar congruences for Gp−1 do exist, but that they are just the particular prime instances of not very well-known but elementary identities for the cycle Stirling numbers. This will lead us to in- troduce a triangular array of integers, the Bernoulli-Stirling numbers, involving the Bernoulli and Stirling numbers which we believe is new.

2. Notation and Preliminaries

In addition to what was exposed in the previous introduction, further notation that we use throughout this paper is presented in this section, along with classical results which we will need. Most of these results can be found in textbooks like [6] and they are given hereafter without proof. In the following, p denotes a prime number, x denotes the argument of a gener- ating function, of a polynomial function or of a formal power series, [[xn]]f(x) the xn m coefficient of n! in f(x) and D f(x) is the m-order derivative of f(x) with respect to x. We will use the Iverson bracket notation: P = 1 when proposition P is true,   and P = 0 otherwise. For q a rational number, we denote by vp(q) the p-adic order of q and we have vp(q1 · q2) = vp(q1) + vp(q2) and vp(q1 + q2) ≥ min (vp(q1), vp(q2)). n
P n
k n The binomial coefficients k are defined by k k x = (1 + x) , whatever the sign of integer n. They obviously vanish when k < 0. They are easily obtained n
n−1
n−1
by the basic recurrence relation k = k + k−1 . When n > 0, we have −n
kn+k−1
k = (−1) n−1 and we have the well-known inversion formula X kn (−1)k−j = [n = j]. (2.1) j k k≥0 n The cycle Stirling numbers k , n ≥ 0 may be defined by the horizontal generating function n−1 X n Y xk = (x + j), (2.2) k k≥0 j=0 where an empty product is meant to be 1. Alternatively, they have the exponential generating function

k
k X nxn (−1) ln(1 − x) = . (2.3) k n! k! n≥0 3

They obviously vanish when k < 0 and k > n. They are easily obtained by the n n−1 n−1 0 basic recurrence k = (n − 1) k + k−1 , valid for n ≥ 1, with k = [k = 0]. They also obey the generalized recurrence relation  n + 1  X h + m n  = . (2.4) m + 1 m h + m h≥0 n We let k , n ≥ 0, be the partition Stirling numbers (or Stirling numbers of the second kind). They also vanish when k < 0 and k > n. Their basic recurrence is n n−1 n−1 0 k = k k + k−1 for n ≥ 1, with k = [k = 0]. They have the following exponential generating function X nxn (ex − 1)k = . (2.5) k n! k! n≥0 We will make use of the following Lemma about the Stirling numbers, for which there is a proof in [6], p. 266-271.

Lemma 2.1. Let n, k be non-negative integers. There exists a polynomial Qk ∈  n  Q[X] of degree 2k, such that Qk(n) coincides with n−k . If k > 0 then 0, 1, ···, k are n+k roots of the polynomial Qk. Moreover, we have Qk(−n) = n and in particular Qk(−1) = 1.

For n ≥ 0, let Bn be the Bernoulli numbers. The first of them are B0 = 1,B1 = 1 1 1 − 2 ,B2 = 6 ,B3 = 0,B4 = − 30 , ... and for h > 0, we have B2h+1 = 0. They have the well-known exponential generating function t X tn = B (2.6) et − 1 n n! n≥0 and they obey the recurrence n X n (−1)nB = B . (2.7) n k k k=0 We will also make use of the Von Staudt-Clausen theorem which states that the denominator of Bn, in reduced form, is the product of all primes p such that p − 1 divides n. In particular, any prime may divide the denominator of a Bernoulli number once at most.

3. Two Identities for the Cycle Stirling Numbers

In this section, we re-demonstrate two identities for the cycle Stirling numbers which, in spite of their similarity to Equation (2.4), do not seem to be very well- known. 4

Theorem 3.1. Let m, n be non-negative integers. We have

 n + 1  X h + m n  = (−1)n−m (−n)h. (3.1) m + 1 m h + m h≥0

Moreover, if n > 0,

 n  X h + m − 1 n  = (−1)n−m (−n)h. (3.2) m m − 1 h + m h≥0

In more compact and symmetric formulations, these two identies also read

 n + 1  X  h n (−1)n−m (−n)m = (−n)h (3.3) m + 1 m h h and, if n > 0,

 n  X  h − 1 n (−1)n−m (−n)m = (−n)h. (3.4) m m − 1 h h

Remark. These identities are not in [6] where quite many finite sums, recurrences and convolutions involving Stirling numbers are reported. Our Equation (3.1) may be obtained as a particular case of Theorem 3 in [2]. An identity equivalent to our Equation (3.2), is obtained incidentally in [1], where it is not even labelled. Another identity, equivalent to our Equation (3.2), is the equation (18) from [15], where it is said to be new.

Proof of Theorem 3.1. Like in [1], our proof will highlight that Equation (3.1) and Equation (3.2) are actually closely related to the convolution Equation (1.1) Qn−1 between the Harmonic and cycle Stirling numbers. Let fn(x) := h=0(x − h). We are going to show that, for m ≥ 1,

m m−1 h n−1 D fn(x) X D fn(x) X 1 m = − (−1)m−h . (3.5) m! h! (x − j)m−h h=0 j=0

Pn−1 Q Pn−1 1 It is true for m = 1, since Dfn(x) = j=0 h6=j(x − h) = fn(x) j=0 x−j . We suppose (induction hypothesis) that it is true for some m, so that

Dm+1f (x) Dmf (x) 1  Dmf (x) (m + 1) n = D n = D m n (m + 1)! m! m m! 5

  m+1 m−1 h+1 n−1 D fn(x) 1 X D fn(x) X 1 (m + 1) = − (−1)m−h (m + 1)! m  h! (x − j)m−h  h=0 j=0   m−1 h n−1 1 X D fn(x) X 1 + (−1)m−h (m − h) m  h! (x − j)m+1−h  h=0 j=0   m h n−1 1 X D fn(x) X 1 = (−1)m−h m  (h − 1)! (x − j)m+1−h  h=1 j=0   m−1 h n−1 1 X D fn(x) X 1 + m (−1)m−h m  h! (x − j)m+1−h  h=0 j=0   m−1 h n−1 1 X D fn(x) X 1 − (−1)m−h m  (h − 1)! (x − j)m+1−h  h=1 j=0   m n−1 1 D fn(x) X 1 = m  (m − 1)! x − j  j=0   m−1 h n−1 1 X D fn(x) X 1 + m (−1)m−h m  h! (x − j)m+1−h  h=0 j=0

m h n−1 X D fn(x) X 1 = (−1)m−h . h! (x − j)m+1−h h=0 j=0 This establishes the validity of Equation (3.5). Now, when x = n, it reads

m m−1 h D fn(n) X D fn(n) m = − (−1)m−h H(m−h). (3.6) m! h! n h=0 This is the same recurrence as in the convolution Equation (1.1), with the same n+1 inital value, since by definition fn(n) = n! = 1 . Then Dmf (n)  n + 1  n = . m! m + 1

Pn n n−h h On the other hand, from Equation (2.2), we have fn(x) = h=0 h (−1) x , and hence n X n h  Dmf (x) = m! (−1)n−hxh−m, n h m h=0 m n    D fn(n) X n h = (−1)n−hnh−m. m! h m h=0 6

Hence n  n + 1  X n h  = (−1)n−hnh−m m + 1 h m h=0 n−m X  n h + m = (−1)n−m (−n)h. h + m m h=0 This completes the proof of Equation (3.1). For the proof of Equation (3.2), we also use an induction argument, but on m and backward. Our induction hypothesis is

n−(m+1)  n  X h + m n  = (−1)n−(m+1) (−n)h. m + 1 m h + m + 1 h=0 Then n−m  n  X h + m − 1 n  = −(−1)n−m (−n)h−1. m + 1 m h + m h=1 Hence n−m  n  X h + m − 1 n  n = (−1)n−m (−n)h. m + 1 m h + m h=1 We subtract the latter equation from Equation (3.1), and we obtain

n−m  n + 1   n  X h + m h + m − 1  n  − n = (−1)n−m − (−n)h. m + 1 m + 1 m m h + m h=1 That is, n−m  n  X h + m − 1 n  = (−1)n−m (−n)h. m m − 1 h + m h=1 To finish the proof, we just need that Equation (3.2) be true for m = n, which is obvious.

(j+1) 4. Extended Congruences for the Harmonic Numbers Gp−1

Theorem 4.1. Let k ≥ 0 be an integer and p a prime number. We have   (k) X j + k (k+j) G = (−1)k (−1)j G pj. (4.1) p−1 j p−1 j≥0 In particular, when k = 0, we have

X j (j+1) j (−1) Gp−1 p = 0. (4.2) j≥0 7

Proof. Letting n = p a prime number, and m = k+1 in Equation (3.2), and dividing throughout by (p − 1)! provides the desired result.

(k) Recall [7] that when k ≥ 1, the generalized Harmonic numbers Hp−1 admit the following p-adically converging expansion:   (k) X j + k − 1 (k+j) H = (−1)k H pj. (4.3) p−1 j p−1 j≥0

It is interesting to point out the similarity of Equation (4.3) and Equation (4.1), but also some differences. Contrary to Equation (4.3), the sum on the right-hand side of Equation (4.1) is finite: it is actually limited to j = p − 1 − k. We also notice that the sign alternates in Equation (4.1) and that there is a slight difference in the binomial coefficient.

It is also known [16] that, for odd prime p,   X j + 2k (2k+j+1) B H (−p)j = 0, (4.4) 2k j p−1 j≥0 the convergence of the series being understood p-adically. More precisely, when p ≥ 5, the following identity was shown in [7]:

2n+1   X j + 2k (2k+j+1) B H (−p)j ≡ 0 (mod p2n+3). (4.5) 2k j p−1 j=0

Now, we look for an equation similar to Equation (4.4), but involving the cycle Stirling numbers. In the case where k = 0, we have, for p ≥ 5

2n+1 X (j+1) j 2n+3 BjHp−1 (−p) ≡ 0 (mod p ). (4.6) j=0

For the first values of n, n = 0, 1, 2..., these congruences read p H + H(2) ≡ 0 (mod p3), p−1 2 p−1 p p2 H + H(2) + H(3) ≡ 0 (mod p5), p−1 2 p−1 6 p−1 p p2 p4 H + H(2) + H(3) − H(5) ≡ 0 (mod p7) ..., respectively. p−1 2 p−1 6 p−1 30 p−1 A clue for our search of an equation analogous to Equation (4.4) involving the Stirling cycle numbers is obtained by making use of Equation (1.1) in order to (j+1) (i+1) recursively compute Hp−1 as function of the Gp−1 , with i ≤ j. Then substituting 8

(j+1) 2n+3 Hp−1 in the above congruences and finally reducing modulo p as much as (i+1) possible, by accounting for any previous congruence involving the Gp−1 . In doing so, the following congruences are found, valid for p ≥ 5:

(2) 3 Gp−1 − pGp−1 ≡ 0 (mod p ), p2 G − pG(2) + G(3) ≡ 0 (mod p5), p−1 p−1 2 p−1 p2 p4 G − pG(2) + G(3) − G(5) ≡ 0 (mod p7), p−1 p−1 2 p−1 6 p−1 p2 p4 p6 G − pG(2) + G(3) − G(5) + G(7) ≡ 0 (mod p9) ... etc. p−1 p−1 2 p−1 6 p−1 6 p−1 The calculations become increasingly laborious as n increases, but we are able to guess that

2n+1 X (j+1) j 2n+3 (j + 1)BjGp−1 p ≡ 0 (mod p ). (4.7) j=0 In an even broader generalization, we anticipate the following theorem which will be proved in the next section. Theorem 4.2. If i ≥ 1 is an integer and p a prime number, then we have   X j + 2i − 1 (j+2i−1) B G pj = 0 (4.8) j j p−1 j≥0 or equivalently,

X j + 2i − 1 p  B pj = 0. (4.9) j j j + 2i j≥0

Remark. Equation (4.8) looks very much like Equation (4.4) but the sums in Equation (4.8) and Equation (4.9) are actually finite.

5. The Aerated Triangular Array An,k

For performing numerical verifications of Equation (4.9), we now introduce the number An,k.

Definition. For non-negative integers n and k, we define the number An,k by

X k + h − 1 n  A := B nh. (5.1) n,k h h h + k h≥0 9

It is clear from this definition that An,k is zero when k > n and that An,n = 1. The first terms of the sequence An,k are computed numerically and displayed in the following table.

n An,0 An,1 An,2 An,3 An,4 An,5 An,6 An,7 An,8 An,9 An,10 0 10 0 0 0000000 1 01 0 0 0000000 2 00 1 0 0000000 3 0 -1 0 1 0 0 0 0 0 0 0 4 0 0 -5 0 1 0 0 0 0 0 0 5 0 24 0 -15 0 1 0 0 0 0 0 6 0 0 238 0 -35 0 1 0 0 0 0 7 0 -3396 0 1281 0 -70 0 1 0 0 0 8 0 0 -51508 0 4977 0 -126 0 1 0 0 9 0 1706112 0 -408700 0 15645 0 -210 0 1 0 10 0 0 35028576 0 -2267320 0 42273 0 -330 0 1

Table 1: The triangular array An,k, for 0 ≤ n, k ≤ 10. It is striking that these numbers seem to be zero when n − k is odd, which, if true, would imply the validity of Theorem 4.2. This will be demonstrated in the next theorem. It is also striking that they seem to be all integers. n−k Theorem 5.1. Let n, k be non-negative integers. We have An,k = (−1) An,k. Equivalently, An,k = 0 when n − k is odd. Proof. When n = 0, this is obviously true. Supposing n > 0, we have X k + h − 1 n  A = B nh n,k h h h + k h     X Bh k + h − 1 n = (−1)n (−1)n−(h+k) (−n)h+k nk h h + k h      X Bh k + h − 1 X g − 1 n = (−1)n (−n)g by Equation (3.4). nk h h + k − 1 g h g k+h−1
g−1
g−k
g−1
But, it is easy to see that h h+k−1 = h k−1 , so that X X g − kg − 1n A = (−1)n−k B (−n)g−k n,k h h k − 1 g g h X g − 1n = (−1)n−k (−1)g−kB (−n)g−k by Equation (2.7) g−k k − 1 g g X k + g − 1 n  = (−1)n−k B ng = (−1)n−kA . g g g + k n,k g 10

Theorem 5.2. For non-negative integers n and k, we have An,k ∈ Z . The proof of this theorem will not be given until the next section. Before proceeding to this proof, we want to point out a similarity between the triangle An,k and the triangle of Stirling numbers of the first kind. We have the following theorem which is analogous to Lemma 2.1.

Theorem 5.3. Let n, k be non-negative integers such that 0 ≤ k ≤ n. There exists a polynomial Pk ∈ Q[X], of degree 2k, such that Pk(n) coincides with An,n−k. Moreover, when k > 0, we have that −1, 0, ···, k are k + 2 roots of Pk(x).

Proof. By definition of An,k, we have

k X n − 1 − (k − h) n  A = B nh, n,n−k h h n − (k − h) h=0 where the binomial coefficient is a polynomial in n from Q[X], of degree h, and  n  after Lemma 2.1, n−(k−h) is a polynomial in n from Q[X], of degree 2(k − h). Therefore An,n−k is also a polynomial in n from Q[X], of degree 2k. Let k > 0,  n  and recall Qj, the polynomial such that Qj(n) = n−j . We have

k X u + h − 1 − k P (u) = B Q (u)uh k h h k−h h=0 k X (u + h − 1 − k) ··(u − k) = B Q (u)uh. h h! k−h h=0 From Lemma 2.1, we know that if k > h then 0, 1, ···, k − h are roots of the polynomial function Qk−h(x). Moreover when k ≥ h, Qk−h(−1) = 1. −1−k
0 Then, when u = 0, we have Pk(0) = B0 0 Qk(0)0 = 0, since k > 0. When u > 0, we have

k X (u − k) ··(u − k + h − 1) P (u) = B Q (u)uh k h h! k−h h=k−u+1 k X (k − u) ··(k − u − h + 1) = (−1)hB Q (u)uh. h h! k−h h=k−u+1

If 0 < u ≤ k, for any h in the set {k −u+1, ··, k} the product (k −u)··(k −u−h+1) must vanish because it has one factor which is zero, and then we also have Pk(u) = 0. 11

Finally, if u = −1,

k X (h − 2 − k) ··(−1 − k) P (−1) = B (−1)hQ (−1) k h h! k−h h=0 k k X (k + 2 − h) ··(k + 1) X k + 1 = B = B h h! h h h=0 h=0 k+1 X k + 1 = B − B = 0 by Equation (2.7). h h k+1 h=0

6. The Dual Triangle Bn,k

We now introduce a dual triangle Bn,k which is similar to the triangle of Stirling numbers of the second kind. Coming back to the polynomial Pk, we may extend the definition of An,k to non-positive indices since, for non-negative n and k, it is natural to define A−n,−n−k := Pk(−n). Then, we have

k X −n − 1 − (k − h) A = B Q (−n)(−n)h −n,−n−k h h k−h h=0 k X n + k = B Q (−n)nh. h h k−h h=0 That is, k−n X k A = B Q (−n)nh −n,−k h h k−n−h h=0 or, accounting for Lemma 2.1,

k−n X kk − h A = B nh. −n,−k h h n h=0

Definition. Let n, k be positive integers. We define the number Bn,k by

X nn − h B := B kh. (6.1) n,k h h k h≥0

It is then clear that for all integers n, k, positive or negative, we have the duality:

A−n,−k = Bk,n. (6.2) 12

−n k This is similar to the duality −k = n which holds [6] for the usual Stirling numbers. It is also clear from this definition that Bn,k is zero when k > n and that Bn,n = 1. Also note that we have Bx+n,x = Pn(−x). The first Bn,k are computed numerically and displayed in the following table.

n Bn,1 Bn,2 Bn,3 Bn,4 Bn,5 Bn,6 Bn,7 Bn,8 Bn,9 Bn,10 Bn,11 Bn,12 1 1000 00000000 2 0100 00000000 3 0010 00000000 4 0 -1 0 1 0 0 0 0 0 0 0 0 5 0 0 -5 0 1 0 0 0 0 0 0 0 6 0 3 0 -15 0 1 0 0 0 0 0 0 7 0 0 49 0 -35 0 1 0 0 0 0 0 8 0 -17 0 357 0 -70 0 1 0 0 0 0 9 0 0 -809 0 1701 0 -126 0 1 0 0 0 10 0 155 0 -13175 0 6195 0 -210 0 1 0 0 11 0 0 20317 0 -120395 0 18711 0 -330 0 1 0 12 0 -2073 0 706893 0 -760100 0 49203 0 -495 0 1

Table 2: The triangular array Bn,k, for 1 ≤ n, k ≤ 12.

Again, we see on Table 2 that the Bn,k seem to be all integers and to vanish when n − k is odd: this will be the next theorem. We can also, as was done in [6] for the usual Stirling numbers, display An,k and B−k,−n in tandem, for n, k ∈ Z. This is the purpose of Table 3, where we have left void the zero entries for k > n and for odd n − k. The numbers which appear now in the diagonal lines are the values of the polynomial function Pn−k(x) for integer arguments.

k \ n -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -8 1 -7 1 -6 -70 1 -5 -35 1 -4 357 -15 1 -3 49 -5 1 -2 -17 3 -1 1 -1 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 2 0 0 0 0 0 1 3 0 0 0 0 -1 1 4 0 0 0 0 0 -5 1 5 0 0 0 0 24 -15 1 6 0 0 0 0 0 238 -35 1 7 0 0 0 0 -3396 1281 -70 1

Table 3: An,k and B−k,−n in tandem, for −8 ≤ n, k ≤ 7. 13

We now proceed to the proof of the integrality of these numbers.

Theorem 6.1. If n and k are non-negative integers, then Bn,k is a triangular array of integers such that Bn,k = 0 when n − k is odd. Moreover, we have the inter-relations X n + xn − x B = A (6.3) x,x−n n − u n + u n+u,u u≥0 X n + xn − x A = B . (6.4) x,x−n n − u n + u n+u,u u≥0 Remark. Theorem 5.2 easily follows from Theorem 6.1. Remark. Equation (6.3) and Equation (6.4) are formally the same as

 x  X n + xn − xu + n = x − n n − u n + u u u≥0  x  X n + xn − xu + n = , x − n n − u n + u u u≥0 respectively, which hold [6] for the usual Stirling numbers.

Proof of Theorem 6.1. For proving that Bn,k is integer, we are going to show that for any prime p, we have vp(Bn,k) ≥ 0. Firstly, we consider the case where p divides k. For all h such that h ≥ 1, we have  n
n−h h vp Bh h k k ≥ 0, since vp(Bh) ≥ −1 by the Von Staudt-Clausen theorem h n
n−0 0
n
and vp(k ) ≥ 1. Moreover, vp B0 0 k k = vp k ≥ 0, obviously. Then n P n
n−h h vp(Bn,k) = vp k + h≥1 Bh h k k ≥ 0. Then, we consider the case where p does not divide k. We may write

X nn − h X nn − h B = B (kh − 1) + B . n,k h h k h h k h≥0 h≥0

By the rule of multiplication of exponential generating functions [17] and given the exponential generating functions Equation (2.3) and Equation (2.6), we have

X nn − h  x (ex − 1)k  B = [[xn]] h h k ex − 1 k! h≥0 1 (ex − 1)k−1  = [[xn−1]] k (k − 1)! 1 n − 1 = . k k − 1 14

 1 n−1  But vp k k−1 ≥ 0, since p does not divide k, and then it suffices to show that h
for all values of h, vp Bh(k − 1) ≥ 0. This is true because either p − 1 does not divide h, and then by the Von Staudt-Clausen theorem, vp(Bh) ≥ 0, or p−1 divides h and then vp(Bh) = −1. But in this case, where p does not divide k and p − 1 h
divides h, Fermat’s little theorem holds and vp k − 1 ≥ 1.

Now, we turn to the proof of the inter-relations Equation (6.3) and Equation (6.4). Let pn be a polynomial of degree n from Q[X]. The set of binomial co- x
efficents k ; 0 ≤ k ≤ n forms a basis for the vector space of the polynomials from Q[X] of degree less than k + 1 and therefore there exists an,k such that Pn x
pn(x) = k=0 an,k k . By the inversion formula Equation (2.1), it is easy to verify P k−uk
that an,k = u(−1) u pn(u). We have seen that Ax,x−n = Pn(x) where Pn is a polynomial of degree 2n from Q[X], so we can apply the above general inversion scheme to Pn(−x) = Bx+n,x and we obtain

2n k X X k −x B = (−1)k−u A x+n,x u u,u−n k k=0 u=0 2n k X X kx + k − 1 = (−1)u A u k u,u−n k=n u=n which, given Theorem 5.1, clearly shows that Bn,k = 0 when n − k is odd. Now, k
x+k−1
x+k−1
x+u−1
since u k = k−u u , we have

2n k X X x + k − 1x + u − 1 B = (−1)u A x+n,x k − u u u,u−n k=n u=n 2n 2n X X x + k − 1x + u − 1 = (−1)u A x + u − 1 u u,u−n u=n k=u 2n X 2n + xx + u − 1 = (−1)u A 2n − u u u,u−n u=n 2n X 2n + x−x = A . 2n − u u u,u−n u=n Then, replacing x by x − n we have

2n X  n + x n − x B = A x,x−n 2n − u u u,u−n u=n n X n + xn − x = A . n − u n + u u+n,u u=0 15

Similarly, let Rn(x) = Bx,x−n, which is a polynomial of degree 2n from Q[X]. We apply the inversion to Rn(−x) = Ax+n,x which gives the similar identity where A and B are exchanged, and this completes the proof of the theorem.

For the usual Stirling numbers, there exist polynomials σn(x) of degree n − 1 from Q[X] (also known as Stirling polynomials in [6], [11]), such that  x  Q (x) = = x(x − 1) ··· (x − n)σ (x) (6.5) n x − n n or equivalently x + n Q (−x) = = (−1)n+1x(x + 1) ··· (x + n)σ (−x). (6.6) n x n

Similarly, accounting for Theorem 5.3, we can define the polynomial Sn(x) of degree n − 2, such that

Pn(x) = Ax,x−n = (x + 1)x(x − 1) ··· (x − n)Sn(x) (6.7) or equivalently

n+2 Pn(−x) = Bx+n,x = (−1) (x − 1)x(x + 1) ··· (x + n)Sn(−x). (6.8)

In the following table, we give the first instances of Sn(x), together with the Stirling polynomial σn(x), given in [6].

n 1 2 3 4

1 1 1 2 1 3 2 σn(x) 2 24 (3x − 1) 48 (x − x) 5760 (15x − 30x + 5x + 2)

1 1 2 Sn(x) 0 − 24 0 5760 (7x + 3x + 2)

Table 4: σn(x) and Sn(x) for n in the range 1 to 4.

The second and third columns of Table 2 are known to the OEIS [14]. Up to the sign and discarding the zeros, we find in these columns the even index Genoc- chi numbers G2n and Glaisher’s G-numbers, A001469 and A002111 at the OEIS, respectively, for which exponential generating functions are known. More generally, we have the following exponential generating function for Bn,k. Theorem 6.2. Let n, k be non-negative integers. We have

X xn (ex − 1)k kx B = . (6.9) n,k n! k! ekx − 1 n≥0 16

Proof. The proof is straightforward: we use the rule of multiplication of exponential generating functions [17] and we have     (ex − 1)k kx X jxj X (kx)j = B k! ekx − 1  k j!   j j!  j≥0 j≥0   X X uh xu = B kj  j h k  u! u≥0 j+h=u   X X uu − j xu = B kj .  j j k  u! u≥0 j≥0

Unfortunately, the derivation of a similar generating function for An,k seems much more difficult.

We finish this section by pointing out a notable difference with the usual Stirling numbers. Whereas it is well-known that the usual Stirling matrices of both kinds 0 are inverses of each other, this is not the case for the An,k and Bn,k. Let An,k be the 0 inverse of the matrix An,k. The first entries of An,k are displayed in the following table.

0 0 0 0 0 0 0 0 0 0 n An,1 An,2 An,3 An,4 An,5 An,6 An,7 An,8 An,9 An,10 1 1 0 0 0 0 0 0 0 0 0 2 0 1 0 0 0 0 0 0 0 0 3 1 0 1 0 0 0 0 0 0 0 4 0 5 0 1 0 0 0 0 0 0 5 −9 0 15 0 1 0 0 0 0 0 6 0 −63 0 35 0 1 0 0 0 0 7 1485 0 −231 0 70 0 1 0 0 0 8 0 18685 0 −567 0 126 0 1 0 0 9 −844757 0 125515 0 −945 0 210 0 1 0 10 0 −14862727 0 600655 0 −693 0 330 0 1

0 Table 5: The triangular array An,k, for 1 ≤ n, k ≤ 10.

0 We do not see any evident link between the matrix An,k and the matrix Bn,k. Moreover, whereas the Stirling matrices are convolution matrices in the sense of [11], this is not the case for the matrices An,k and Bn,k. This is easily checked on their first entries, as indicated in [11]. If An,k and Bn,k were convolution matrices, P xn f(x)k inverse to each other, there would exist a function f(x) such that n Bn,k n! = k! P xn g(x)k and the generating function for An,k would read n An,k n! = k! , where g is the 17

n! compositional inverse of f. The triangular array k! Bn,k is not even a Riordan array for which (see for instance [13]) the ordinary generating function reads d(x) · h(x)k for some power series d(x) and h(x), with h(0) = 0 and Dh(0) 6= 0. Knowing the generating function Equation (6.9) for Bn,k, the difficulty to find an analogous generating function for An,k has to do with these observations.

7. Discussion and Questions

Since the second column of the triangle Bn,k corresponds to the Genocchi numbers, we might consider the other columns as some sort of generalized Genocchi numbers. However, these numbers are not the same as the already known generalized Genocchi numbers from [4], nor as those from [12]. For the classical Genocchi numbers, there exists a recursion so that for n ≥ 1, it is possible to compute B2n,2 recursively:

n−1 1 X 2n B = n − B . 2n,2 2 2j 2j,2 j=1

Moreover, a combinatorial interpretation has been given to the Genocchi numbers [5], but to the author’s knowledge, this is not the case for Glaisher’s G-numbers, |B2n+1,3|. Here we raise the more general questions: for a given k > 2, find a recursion for the Bernoulli-Stirling numbers of the second kind Bn,k and find com- binatorial objects that they enumerate.

As for the Bernoulli-Stirling numbers of the first kind An,k, we have even more questions. Apart from their appearance in the above investigation of congruences modulo prime powers for the cycle Stirling numbers, we don’t know their math- ematical interest. Any recurrence that would allow to compute an entry in this triangular array from entries of previous lines would be insightful, and might lead to a direct proof of Theorem 5.2. Moreover An,k cries for a generating function, of any kind, or at least a functional equation involving such a function. There also remains the problem of the combinatorial interpretation of An,k.

By comparison to these quite complicated combinatorics questions, the study of the arthmetic properties of the Bernoulli-Stirling numbers would seem more easy, as it could be made use of their explicit expression in terms of Bernoulli and Stirling numbers and then take advantage the existing knowledge on the arithmetic proper- ties of the latter. In particular, we might expect that the Bernoulli-Stirling numbers satisfy some sort of Kummer congruence, as do the regular Bernoulli numbers. 18

References

[1] V. Adamchik, On Stirling numbers and Euler sums, J. Comput. Appl. Math. 79 (1997), 119-130.

[2] T. Agoh and K. Dilcher, Convolution identities for Stirling numbers of the first kind, Integers 10 (2010), 101-109.

[3] L. Carlitz, Some partition problems related to the Stirling numbers of the second kind, Acta Arith. 10.4 (1965), 409-422.

[4] M. Domaratzki, Combinatorial interpretations of a generalization of the Genocchi numbers, J. Integer Seq. 7 (2004), Article 04.3.6.

[5] D. Dumont, Interpr´etationscombinatoires des nombres de Genocchi, Duke Math. J. 41 (1974), 305-318.

[6] R.L. Graham, D.E. Knuth and O. Patashnik, Concrete Mathematics, 2nd Edition, Adison- Wesley Publishing Company, 1994.

[7] R. Gy, Extended congruences for certain harmonic numbers (2019), appears on arXiv as https://arxiv.org/abs/1902.05258.

[8] F.T. Howard, Congruences for the Stirling numbers and associated Stirling numbers, Acta Arith. 55 (1990), 29-41.

[9] D. Kalman. A matrix proof of Newton’s identities, Math. Mag. 73 4 (2000), 313-315.

[10] J. Katriel, A multitude of expressions for the Stirling numbers of the first kind, Integers 10 (2010), 273-297.

[11] D.E. Knuth, Convolution polynomials (1992), appears on arXiv as https://arxiv.org/abs/math/9207221v1.

[12] Q.M. Luo and H.M. Srivastava, Some generalizations of the Apostol-Genocchi polynomials and the Stirling numbers of the second kind, Appl. Math. Comput. 217 (2011), 5702-5728.

[13] A. Luz´on,D. Merlini, M.A. Mor´on,R. Sprugnoli, Identities induced by Riordan arrays, Linear Algebra Appl. 436 (2012), 631-647.

[14] N.J.A. Sloane, The On-Line Encyclopedia of Integer Sequences, published electronically at https://oeis.org.

[15] V. Shevelev, On identities generated by compositions of positive integers, (2012), appears on arXiv as https://arxiv.org/abs/1211.1606.

[16] L.C. Washington, p-adic L-functions and sums of powers, J. Number Theory 69 (1998), 50-61.

[17] H.S. Wilf, Generatingfunctionology, 2nd Edition, Academic Press, 1992.